Question

7. 25 ml of 6 M NaOH was diluted with water to 400 ml. What is the pH and pOH of the final solution? What is the [H+] and [OH-]? (Remain calm; remember the formula for dilution from the previous lab session: (V1)(C1) = (V2)(C2) )

8. A solution was made by dissolving 10 g of NaOH (formula weight 40 g/mole) in water and the final volume brought to 500 ml. What is the pH, pOH, [H+] and [OH–] of this solution? I have no idea how to work these problems out and would like help. So please if anyone could answer these problems as soon as possible that would be great.

Answer 1

Q7.

mmol of NaOH = Mbase*Vbase = 25*6 = 150 mmol of NAOH

dilution goes to V = 400 mL

[NaOH]diluted = mmol/mL =  150 /400 = 0.375 M

[OH-] = [NaOH] = 0.375 M

pOH = -log([OH-])

pOH = -log(0.375 )

pOH =0.4259

pH = 14-pOH = 14-0.4259

pH = 13.5741

[H+] = 10^-pH = 10^-13.5741 = 2.6662*10^-14 M H+

Q8.

mol of NAOH = mass/MW = 10/40 = 0.25 mol of NaOH

[NaOH] = mol/V = 0.25 / 0.5 = 0.5 M

[OH-] = 0.5 M

pOH = -log(OH) = -log(0.5) = 0.3

pOH = 0.3

pH = 14-pOH = 14-0.3 = 13.7

pH = 13.7

[H+] = 10^-pH = 10^-13.7

[H+]= 1.995*10^-14 M